Problem: $\text E = \left[\begin{array}{rr}-1 & -2 \\ -2 & -1\end{array}\right]$ and $\text A = \left[\begin{array}{rr}2 & 0 \\ 2 & -1\end{array}\right]$. Let $\text {H = EA}$. Find $\text H$. $ {H = }$
Solution: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{E}$ and the first column of $\text{A}$. $ \text {H}=\left[\begin{array}{rr}{-1} & {-2} \\ -2 & -1\end{array}\right]\left[\begin{array}{rr} {2} & 0 \\ {2} & -1\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(-1,-2)\cdot(2,2)\\\\ &=-1 \cdot 2 - 2\cdot 2\\\\ &=-6 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-1 \cdot 0 - 2\cdot 1 = -2$ (Choice B) B $-2 \cdot 2 - 1\cdot 2 = -6$ (Choice C) C $-2 \cdot 2 - 1\cdot 0 = -4$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}-6 & 2 \\ -6 & 1\end{array}\right]$